The frequency of the first harmonic of a string stretched between two points is 100 Hz . The frequency of the third overtone is

Options

A. `200 Hz`
B. `300 Hz`
C. `400 Hz`
D. `600 Hz`

Answer

To find the frequency of the third overtone given that the frequency of the first harmonic is 100 Hz, we can follow these steps:

### Step 1: Understand the relationship between harmonics and frequency
The frequency of the nth harmonic of a string fixed at both ends can be expressed as:
\[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f_n \) is the frequency of the nth harmonic,
- \( n \) is the harmonic number (1 for the first harmonic, 2 for the second harmonic, etc.),
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Identify the given information
From the problem, we know:
- The frequency of the first harmonic (n = 1) is 100 Hz.

### Step 3: Determine the frequency of the third overtone
The third overtone corresponds to the 4th harmonic (since the overtone numbering starts from the first harmonic):
- First harmonic (n = 1)
- Second harmonic (n = 2)
- First overtone (n = 3)
- Second overtone (n = 4)
- Third overtone (n = 5)

Thus, for the third overtone, we have:
- \( n = 5 \)

### Step 4: Calculate the frequency of the third overtone
Using the relationship for frequency:
\[ f_5 = \frac{5}{2L} \sqrt{\frac{T}{\mu}} \]

Since we know that:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = 100 \text{ Hz} \]

We can express \( f_5 \) in terms of \( f_1 \):
\[ f_5 = 5 \times f_1 \]
\[ f_5 = 5 \times 100 \text{ Hz} = 500 \text{ Hz} \]

### Final Answer:
The frequency of the third overtone is **500 Hz**.

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